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Viscosity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Viscosity is considered one of the most asked concept.

  • 23 Questions around this concept.

Solve by difficulty

A beaker contains a fluid of density \rho kg/m3, specific heat S J/kg 0C and viscosity \eta. The beaker is filled up to height h. To estimate the rate of heat transfer per unit area (Q /A) by convection when a beaker is put on a hot plate, a student proposes that it should depend on \eta,

\left ( \frac{S\Delta \Theta }{h} \right ) \: and\: \left ( \frac{1}{\rho g} \right )when \Delta \Theta   (in_{}^{0}\textrm{C}) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for (Q /A) is:

 

Match List I with List II       

                    List I                         List II
  A. Surface tension        \text { I. } \mathrm{kgm}^{-1} \mathrm{~s}^{-1}
  B. Pressure        \text { II. } \mathrm{kgms}^{-1}
  C. Viscosity        \text { III. } \mathrm{kgm}^{-1} \mathrm{~s}^{-2}
  D. Impulse         \text { IV. } \mathrm{kgs}^{-2}

Choose the correct answer from the options given below:


                                          


 

 

 

 

 

 

 

Match List I with List II

                           List I                               List II
A. Young's Modulus \left ( Y \right )                    I .                      \left [ ML^{-1} T^{-1} \right ]        
B. Coefficient of Viscosity \left ( \eta \right ) II.  \left [ ML^{2}T^{-1} \right ]
C. Planck's Constant \left ( h \right ) III.  \left [ ML^{-1}T^{-2} \right ]
D. Work Function \left ( \phi \right ) IV.  \left [ ML^{2}T^{-2} \right ]

Choose the correct answer from the options given below: options

Match List I with List II

  List-I

LIST – II

A. Torque I.

\mathrm{ML^{-2}T^{-2}}

B. Stress II.

\mathrm{ML^{2}T^{-2}}

C. Pressure gradient III. \mathrm{ML^{-1}t^{-1}}
D. Coefficient of viscosity IV \mathrm{ML^{-1}T^{-2}}

Choose the correct answer from the options given below:

Choose the correct relation about Newton's law of viscosity -

As the temperature of water increases, its viscosity.

 

A small steel ball falls through a syrup at a constant speed of 1 \mathrm{~m} / \mathrm{s}. If the steel ball is pulled upward with a force equal to twice its effective weight, how fast will it move upward?
 

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A large wooden plate of area \small 10 \mathrm{~m}^2 floating on the surface of a river is made to move horizontally with a speed of \small 2 \mathrm{~m} / \mathrm{s} by applying a tangential force. The river is \small 1 \mathrm{~m} deep and the water in contact with the bed is stationary. Then choose the correct statement-

\small \text { Choose the correct option. }

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Concepts Covered - 1

Viscosity
  • Viscosity-

 Ideal fluids are non-viscous. But for real fluids, there is a viscous force between the adjacent layers of fluids which are in contact.

In case of a steady flow of a fluid when a layer of fluid slips or tends to slip on adjacent layers in contact, the tangential force/viscous force acting between two adjacent layers try to stop the relative motion between them.

So The property of a fluid due to which it opposes the relative motion between its different layers is called viscosity. 

Viscosity is also known as fluid friction or internal friction.

  • Velocity gradient -

It is defined as the ratio of change in velocity to change in height.

I.e Velocity gradient $=\frac{\text { chane in velocity }}{\text { change in height }}$

For the above figure

Layer AB is at rest

While Layer CD is having velocity v and is at a distance x from layer AB

Similarly, Layer MN is having velocity (v+dv) and is at a distance (x+dx)  from layer AB

Then Velocity gradient is given as Velocity gradient $=\frac{d v}{d x}$
Means Velocity gradient denotes the rate of change of velocity with distance x .
- Viscous Force-

In case of a steady flow of fluid, the force between the fluid layers opposing the relative motion is called viscous force.
1. Viscous force directly proportional to the area
l.e $F \alpha A$

Where A is the area
2. Viscous force directly proportional to the Velocity gradient

$
\text { I.e } F \alpha \frac{d v}{d x}
$


So we can write

$
F \alpha \frac{A d v}{d x}=F=-\eta A \frac{d v}{d x}
$


Where $A-$ Area
F-Viscousforce
$\eta=C o-e f f$ ficient of viscosity
$v-$ Velocity of liquid
$x$ - Distance from refrence point

And here Negative sign shows viscous force acts opposite to flow of liquid

  • Coefficient of viscosity-

    From the equation
    $
    F=-\eta A \frac{d v}{d x}
    $

    $
    \begin{aligned}
    & \quad \text { If } A=1 \text { and } \frac{d v}{d x}=1 \\
    & \text { then } \eta=F
    \end{aligned}
    $


    So the coefficient of viscosity is defined as the viscous force acting per unit area between two layers moving with unit velocity gradient.
    1. Coefficient of viscosity shows the nature of liquids.
    2. Unit of viscosity is dyne $-s-\mathrm{cm}^{-2}$ or Poise in CGS system

    And Newton $-S-m^{-2}$ or Poiseuille or decapoise ${ }_{\text {in the SI system }}$
    And 1 decapoise $=10$ Poise
    3. The dimension of viscosity is $M L^{-1} T^{-1}$

  1. The cause of viscosity in liquids is cohesive forces among molecule whereas, in gases, it is due to the diffusion of molecules.

  2. The viscosity of the liquid is much greater (about 100 times more) than that of gases.

  3. With an increase in pressure, the viscosity of liquids (except water) increases while For gases viscosity is practically independent of pressure. And the viscosity of water decreases with increase in pressure.

  4. The viscosity of gases increases with the increase of temperature, because on increasing temperature the rate of diffusion increases.

  5. The viscosity of liquid decreases with the increase of temperature, because the cohesive force between the liquid molecules decreases with the increase of temperature

 

  • Poiseuille’s Formula

For the stream-line flow of liquid in capillary/narrow tube, If a pressure difference (P) is maintained across the two ends of a capillary tube of length 'l ' and radius r as shown in figure

Then according to Poiseuille’s Formula

V= the volume of liquid coming out of the tube per second is

  1. Directly proportional to the pressure difference (P).

  2. Directly proportional to the fourth power of radius (r) of the capillary tube

  3. Inversely proportional to the coefficient of viscosity (\eta) of the liquid.

  4. Inversely proportional to the length (l) of the capillary tube.

And with the help of Dimension formula we get

$
V=\frac{K P r^4}{\eta l}
$


Where K is the constant of proportionality
And experimentally it is found that $K=\frac{\pi}{8}$
So Poiseuille's Formula is given as

$
V=\frac{\pi P r^4}{8 \eta l}
$
 

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Viscosity

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Viscosity

Physics Part II Textbook for Class XI

Page No. : 262

Line : 47

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